@MikeMiller @BalarkaSen do you like my suggestion? (on the starboard)

+1

please star it lol

also, do you think we need more active owners?

maybe we can make the more active users in this room owner

how to do a +- below simulataneously in latex any idea?

@BAYMAX \pm

@BAYMAX it's not really a latex channel

I have no opinions in that matter. If the room feels like it, I have no problem with more owners.

@BalarkaSen detexify.kirelabs.org/classify.html

@felipa wrong ping?

sorry

I feel this room is a friends room and i ask my friends

@LeakyNun it was for help finding latex symbols in general

@BalarkaSen do you think I should ask it on meta?

Alright, I've had some Advil and stuff, hello chat.

@Leaky No idea

@Fargle Rehi!

@LeakyNun If you do that, don't make me owner. I'll just ban nerds.

How goes it @Balarka?

I'm blasting Fear of Music into headphones

and learning about simplicial sets lol

What would the advantages of a new room owner be?

An incredible album.

I will find a city, find myself a city to live in

But I definitely found a band to listen to :P

@SteamyRoot if the room owners are more active then it would certainly bring about advantages

Like...?

If you ever can, try to get a hold of their film Stop Making Sense.

@SteamyRoot no idea at all

@SteamyRoot Eg less flagging dystopia

There would be an active guy to dismiss bad flags

But eh, I think we're stable as it is

Can a room owner do that?

@Fargle Oh, neat, I will.

I think? No idea.

Afaik only moderators can do that.

I think 10k users can dismiss flags

It's a great film, mostly because David Byrne wears a massive suit that makes his head look tiny, and it gets progressively bigger over the course of the film.

lol

Vexillologists can dismiss flags

Keksillologists

logologists

I'll be back shortly, off to get utterly unhealthy breakfast from a corporation with no stake in my well-being.

:(

Bon apetit, maybe

Hello, please is this differential equation : y y''+y'=x nonlinear and how we can solve it ?

@Vrouvrou ask wolframalpha

wolframalpha.com/input/?i=y+y%27%27%2By%27%3Dx

we can't solve it ?

@Vrouvrou then we can't

You're saying we can't because wolfram can't?

yes i don't know how to do

@Vrouvrou Solving second-order non-linear ODE's tends to be really difficult, unless they're of a specific form.

Was this an exercise, or something you came up with yourself? Because, in general, an analytic/closed form solution need not exist.

no i just found it as an example of a nonlinear D.E

i don't know why it is a nonlinear D.E

It has a term $yy''$

we put $f(y)=yy''$ and we see if $f(ay+z)=af(y)+f(z)$ ?

You should check that $f(ay_1 + by_2) = af(y_1) + bf(y_2)$ for $f(y) = yy'' + y'$.

ok

thank you

@Fargle Fun fact: "Breakfast is the most important meal of the day" has never been backed up by scientific evidence

It was originally a corporate slogan from, I think Kellogg's?

@AkivaWeinberger I'd heard this before.

Huh, apparently if $A$ is an $n\times n$ matrix, then $A$ can't have order greater than $n$

If $A^{n+1}=0$, then $A^n=0$

@AkivaWeinberger cayley hamilton

If $A$ has finite order, you mean?

(This was probably a Tedcercize I forgot about or something)

@BalarkaSen Yeah, sorry

Wait, actually, I might have used the word wrong.

Is order the smallest $m$ where $A^m=0$ or the smallest $m$ where $A^m=I$?

Yeah it's the latter.

Oh. Whoops.

Maybe call it nilpotency order.

Well, in any case, if $A^{n+1}=0$, then $A^n=0$.

@LeakyNun I don't see how that would prove it

Nilpotent implies it has characteristic polynomial x^n doesn't it?

@AkivaWeinberger its characteristic equation must be disivisible by a power of $A$

So by C-H, A^n = 0.

wait, that doesn't prove it

@BalarkaSen I don't know how to prove that

Eigenvalues. The only eigenvalue of A is 0, with multiplicity n.

@BalarkaSen Oh. I see. Cool

That's a completely different proof

The one I just saw constructed $n+1$ linearly independent vectors on the assumption $A^n\ne A^{n+1}=0$

oh wait

CH implies that $\{A^r \mid r \le n\}$ is linearly dependent

right

(The proof I saw didn't use C-H, as well, for what it's worth)

so you can express $A^n$ in terms of smaller powers

then $A^{n+1} = A^n A = P(A) A$

with degree $n$

then use CH again

You could take the characteristic polynomial and multiply it by a power of $x$ such that the lowest degree term becomes $cx^n$

Plugging in $A$ gives $cA^n=0$ (everything above vanishes)

I should learn linear algebra someday.

Their proof was,

if $A^n\ne0$, then $A^nu\ne0$ for some vector $u$.

They then prove that $\{u,Au,A^2u,\dots,A^nu\}$ are linearly independent, contradiction.

Kewl.

I like it.

They're linearly independent, because if $c_0u+c_1Au+\dotsb+c_nA^nu=0$,

we can prove $c_0=0$ by multiplying it by $A^n$ (everything above vanishes)

Once we have that, we show $c_1=0$ by multiplying by $A^{n-1}$.

@AkivaWeinberger how does it fail if I replace $n$ with $n-1$?

How exactly are parabolic subgroups of real semisimple Lie groups defined? Is it related to the usual definition of a parabolic subgroup of a linear algebraic group?

@aminliverpool This is known as Hilbert's Lemma.

Suppose the point is not umbilic. Take a suitable parametrisation such that $k_1 = L/E$ and $k_2 = N/G$. Use Codazzi to find the first and second derivatives of $k_1$ and $k_2$

And use that to prove $K$ must then be negative.

yup

principal directions are orthogonal

so choose a parametrisation $x(u,v)$ with $x_u$ in the first principal direction and $x_v$ in the second principal direction. Then $x_u$ and $x_v$ are orthogonal

Principal directions are eigenvectors of the shape operator

Ummm... You pick your parametrisation so the coordinate lines lie in the principal directions at the point $p$

Ummm... any non-umbilic point allows such parametrisation. See for example math.stackexchange.com/questions/1059905/…

Here, then. math.stackexchange.com/questions/1166858/…

What is the (topological) join of the integers with itself? $\Bbb Z * \Bbb Z$?

What do you mean by the topological join?

$X \times Y \times I$ modulo pinching the ends $X \times Y \times i$ for $i = 0, 1$ to $X$ and $Y$ respectively?

$X* Y = X\times Y\times [0,1]/ rel$ where $X\times\{y\}\times\{0\}$ is identified to a point and $\{x\}\times Y\times\{1\}$ is too for any $x$ and $y$

yes

im drawing a picture but it looks really bad

I don't think there's a very good description

for $\Bbb Z_2 * \Bbb Z_2$ you can see that you get $S^2$

Also, what a coincidence, I have been thinking about the bar construction lately

where are the ends of $\Bbb Z$?

(which involves iterated join of groups)

@Balarka the infinite join should be contractible and somehow "nicely related" to $\Bbb R$ (beyond homotopy equiv. i think)

whats the bar construction?

@LeakyNun $\Bbb Z$ never ends

@s.harp my intuition isn't working for this one lol

@s.harp Err? Z/2 x Z/2 x I is like 4 parallel lines, isn't it? Z/2 * Z/2 should be S^1.

@LeakyNun draw the lattice $\Bbb Z^2$, draw lines going up so that all points that have the same $x$ value meet and line going down so that all points that have hte same $y$ avlaue meet

@Balarka oops thats what i meant

@s.harp nope. mind=blown. never mind.

@s.harp It's a description of the classification space of a topological group $G$ as a "limit" of $G * G * \cdots * G$

($n$ times) where $n \to \infty$

@Balarka if you take mod the $G$ action you get the classifying space ;)

I know that as the milnor construction, which I am currently reading through for a seminar talk on thursday

Correct.

@BalarkaSen you should learn some linear algebra

This is crazy, we are reading about the same things.

Wanna learn with me?

im using the book by Husemoller, are you using any special literature?

sure^

@s.harp Ya, the bar construction is very closely related to it.

@Leaky I agree.

I am not reading a book particularly, just picking stuff up from a bunch of notes the author sent me via email. If you want I can email it to you.

@aminliverpool I already mentioned that before

$x_u$ and $x_v$ are orthogonal if they point in the principal directions.

that would be nice

Got it.

@LeakyNun what?

hi chat

Why are you trolling, @LeakyNun?

I just wanted to say that you shouldn't post your email here lol

ah^

but you removed it already @s.harp

@BalarkaSen conveying a message in a non-standard manner is not trolling

I thought you were saying that it auto-censored the email address

morning chat

chat is lagging here

Why are you conveying a message in a nonstandard manner?

Speak clearly so that ye may be understood

Something I thought I'd never have to read up on again: The Foucault pendulum

@aminliverpool they are the eigenvalues of the shape operator. Either take a good look at the shape operator or just calculate them?

@Semi the book by Umberto Eco or the pendulum itself? :D

Pendulum itself.

@BalarkaSen never mind

Because that's apparently connected to geometric phase somehow.

@LeakyNun I didn't. I was just asking a question :)

Yes, but these aren't any eigenvectors.

What properties does the shape operator obey? Typically that's enough to tell you about the eigenvectors.

@s.harp Did the mail go? It tells me it didn't.

Look, seriously, if you have to ask all these questions, either you're not trying and just want me to write down a full proof for you; or you're clearly not ready to prove Hilbert's lemma. Whatever it may be, I'm out, I have work to do of my own...

@leakynun I'll bite. What's the meaning of the proposed Greek?

@Semiclassical ...let's just agree that I was trolling.

do whatever you want

oook

@balarka it appears I mixed up two email addresses, replace the @kip.u... with @stud.u...

Ah ok

done.

ok I have recieved it, thanks!

No problem.

I am not actually sure if it's obvious to describe $B \Bbb Z/2$ as $\Bbb{RP}^\infty$ from the Milnor construction.

you can explicitly describe $EG$ (the infinite join) and its topology, there you can see that its the same as $S^\infty$ and the action of $\Bbb Z/2$ is the the same as $x\mapsto -x$

namely the points of EG are of the form $(t_1 x_1,...,t_n x_n,0,....,0)$ and the open sets can be described the ones that make the maps $t_i: EG\to [0,1]$, $x_i: t_i^{-1}(0,1]\to G$ continuous

if you do that with $\Bbb Z_2$ you see that the points are the same as points in $\Bbb R^\infty$ ($c_{00}(\Bbb N)$ with $1$-norm to be precise) that have norm $1$

Hmm

Oh, I see. $\Bbb Z/2 * \Bbb Z/2 * \Bbb Z/2 = S^1 * \Bbb Z/2$, which is like $S^1 \times S^0 \times [0, 1]$ with in the $S^1 \times S^0 \times 0$ end, the two circles identified, and in the $S^1 \times S^0 \times 1$ end, the two circles pinched.

yes^

That's a cylinder with the two ends pinched, I guess. So an $S^2$.

I guess you can show $(\Bbb Z/2)^n$ is $S^n$. Then you're just taking $\text{lim}_{n \to \infty} S^n$ (under an appropriate direct system).

Got it, thanks

and the action of $\Bbb Z/2$ on this guy switches the $\Bbb Z/2$ components and applies the action of $\Bbb Z/2$ on the $S^{n-1}$ componennt

Ahhh

Right, so you're just making it $\Bbb{RP}^n$ at each skeleton, and taking limit.

you can really draw a picture

yes

This is cool. Neato.

but im not so well versed with categorical notions to know whether or not a limit of projections gives the projection on the limit

Maybe I can answer that depending on what you mean by projection. There's this theorem which says that if $X_1 \to X_2 \to X_3 \to \cdots$ is a directed system and $f_i : Y \to X_i$ are maps which commute with the system (that is $d_i \circ f_i = f_{i+1}$, where $d_i : X_i \to X_{i+1}$ are the bonding maps), then there is a natural map $f : Y \to \varinjlim X_i$ which extend the $f_i$'s.

Similar for maps $g_i : X_i \to Y$ to $Y$.

(That's basically all I know about the category theory of limits lol)

Ok, I guess you meant the covering projections $S^n \to \Bbb{RP}^n$ at each stage. Yeah, this should definitely give a map $S^\infty \to \Bbb{RP}^\infty$ under the "easy" directed system (each $S^i$ considered as an equator of $S^{i+1}$). Should be rigorously doable by the naturality theorem I described.

ah

yes this is what I was thinking about

The picture gets somewhat ugly for $B\Bbb Z$.

Maybe $\Bbb Z * \Bbb Z$ is homeomorphic to $\Bbb R$, though.

it cant be, because if you look at the picture i sketched above ($\Bbb Z^2\times I$, connect all points with same $x$ value at $1$ and all points with same $y$ value at $0$) you have points that look like a "fan" $(\bigcup_n I_n) /\{0_n\equiv 0_m\}$

Oh right sorry, yes.

Ok, this is beyond ugly

its countably many of those fans in a row on the top and countably many at the bottom in a row going prepedicular to the previous row so that each pair has exactly one line between them

guys any thoughts on this ? math.stackexchange.com/questions/2337725/…

ok im going to a seminar, be back in 2 hours

Hi chat

Are you still here @Balarka?

Yeah

Does that ring a bell to you?

I remember either you or Mike quoting a result according to which a big class of topological spaces is in some sense equivalent to finite spaces, something like "every (insert conditions here) is homotopy equivalent to a finite space" but I can't find it nor remember the details

Ah yes, finite simplicial complexes.

Not homotopy equivalent to; weak homotopy equivalent to.

(That means there is a map $f : X \to Y$ which induces isomorphism in homotopy groups of all dimensions)

Ah, I see, thanks!

I think this is called finite homotopy theory.

here

Does the weak in the name suggest that 2 spaces can be weak homotopy equivalent without being homotopy equivalent?

Yes. But if X, Y are CW complexes, weak homotopy equivalent <=> homotopy equivalent.

Interesting, I'll think about an example where the implication works in only one direction

You need to come up with a sufficiently bad space.

I have to go already (I'm in a part of Tuscany with no internet and bad phone coverage now), thanks for your help

No problem, see ya.

@BalarkaSen Question. How much do you know re: Lagrangian and/or Hamiltonian mechanics?

Not much, actually.

I just know they are things :P

lol

I'm trying to decide whether I should learn the differential geometry versions of those, mostly so that I could properly understand some of this geometric phase business

Sounds like a question for Ted.

yeah.

My advisor gave me a question to think about and I'm pretty sure it doesn't actually make sense in retrospect.

So you know that you are given a physics question.

In that I can push it in two different directions. In one direction the idea definitely doesn't make sense, and in the other I think it's pretty much already known.

Well, have you heard of Foucault's pendulum?

Nope. I should, I guess.

Yeah, it's a pretty classic experiment.

Oh, the one which demonstrates the rotation of earth.

Right.

I do know of that one.

Apparently there's a connection between that and this geometric phase business.

@Secret playing with math..

One version of the Foucault pendulum idea would be this: Start a spherical pendulum in a circular rotation rather than a linear oscillation.

@MikeMiller I recently read a theorem which says given a "generic" (what does that mean?) Morse function $f$ on a smooth manifold $M$, $BC_f \cong M$ where $C_f$ is the category whose objects are critical points of $f$ and maps between objects are gradient flows lines between the critical points.

Pretty fantastic.

In that case, the effect of the Earth's rotation would be that there's a phase difference of $2\pi\cos\theta$ between the pendulum's initial circular rotation and its circular rotation after going around the earth once.

($\theta$ being the angle in spherical coordinates relative to the earth's rotational axis)

To get the usual Foucault pendulum, one just notes that linear oscillation is a superposition of two opposite circular motions. (i'm stealing this from a Michael Berry article)

but I'm not saying anything useful. blarg.

There's somehow a link between this and the concept of parallel transport, is the thing.

From Wikipedia:

"Rather than tracking the change of momentum, the precession of the oscillation plane can efficiently be described as a case of parallel transport. For that, it can be demonstrated, by composing the infinitesimal rotations, that the precession rate is proportional to the projection of the angular velocity of Earth onto the normal direction to Earth, which implies that the trace of the plane of oscillation will undergo parallel transport."

After 24 hours, the difference between initial and final orientations of the trace in the Earth frame is $\alpha = −2\pi \sin \phi$, which corresponds to the value given by the Gauss–Bonnet theorem. $\alpha$ is also called the holonomy or geometric phase of the pendulum. When analyzing earthbound motions, the Earth frame is not an inertial frame, but rotates about the local vertical at an effective rate of $2\pi \sin \phi$ radians per day.

($\phi$ here is the latitude.)

@BalarkaSen This is usually called a flow category (I would actually anticipate that you have to include positive-dimensional spaces of trajectories in the morphisms too)

this is a pretty productive idea but I think hard to get a lot of relevant technical details to work out

@Semiclassical Oh, strange.

@MikeMiller ah ok.

Yeah. So somehow there's some neat math involved.

didn't know the name

The article here looks like it gives a readable treatment of it: maa.org/sites/default/files/pdf/upload_library/22/Ford/…

@MikeM The example in the notes give the height function of a tilted torus as an example, and claims the space of flowlines from the index 1 critical point to the index -1 is 4 dimensional. I see 2 gradient flowlines, are the other two "going across" the donut-hole?

I don't think I see them as clearly.

can anyone see how fermats little theorem can solve y^3 + z^3 ≡ 0 (mod 503) note that 503 is a prime

@AliceRyhl Shot in the dark: Since 503 is prime, every number from 1 to 502 has a multiplicative inverse.

So if y isn't zero mod 503 then y^-1 must exist.

so y^-3 z^3 ≡ -1 (mod 503)

Right. Or, (zy^-1)^3=-1 mod 503.

so you're trying to solve x^3 = -1 mod 503.

Which seems a lot more Fermat than what you started with.

yeah

Of course, you should also handle the case of y=0 mod 503. But that's pretty trivial.

So x^6 = 1 mod 503. But you also know x^502 = 1 mod 503, right?

What does that tell you?

x^503 = (x^6)^83 * x^4 = 1 mod 503 so x^4 = 1 mod 503 and similarily x^2 = 1 mod 503

Right. But you already had x^3=-1 mod 503.

x^4 = 1 and x^3 = -1 gives x = -1 mod 503

Right.

So what can we conclude about y and z?

so zy^-1 = -1 mod 503 which is z = -y mod 503